Question

Three numbers are chosen from 1 to 15 . Find the probability that they are consecutive. (1) \(1 / 35\) (2) \(1 / 13\) (3) \(2 / 13\) (4) \(3 / 35\)

Short answer

Answer: \(1/35\)

Step-by-step solution

Calculate the total number of possible cases

To choose any three numbers from 1 to 15, we can use the combination formula C(n, r) = n! / [r!(n-r)!], where n is the total number of elements and r is the number of elements we need to choose. In this case, we have n = 15 (numbers 1 to 15) and r = 3 (choosing three numbers). By plugging these values into the combination formula: Total possible cases: C(15, 3) = 15! / [3!(15-3)!] = 455 There are 455 possible cases when choosing any three numbers from 1 to 15.

Calculate the number of favorable cases (Consecutive numbers)

To count the number of favorable cases, we can simply go through the list of consecutive numbers and see how many sets of three consecutive numbers are there: 1, 2, 3; 2, 3, 4; 3, 4, 5; ...; 12, 13, 14; 13, 14, 15 We can count the sets or identify a pattern: in the first set, the smallest number is 1. In the second set, the smallest number is 2. This pattern continues until the last set, where the smallest number is 13. We have a total of 13 favorable cases, where three consecutive numbers are chosen.

Calculate the probability

Now that we know the number of favorable cases and the total number of possible cases, we can calculate the probability by dividing the number of favorable cases by the total number of possible cases: Probability of choosing three consecutive numbers = Number of favorable cases / Total possible cases Probability = 13 (favorable cases) / 455 (total possible cases) The probability turns out to be (13/455). If we simplify this fraction, we obtain: (13/455) = (1/35) Now, let's compare this result to the given options: (1) \(1 / 35\) <- This is our calculated probability. (2) \(1 / 13\) (3) \(2 / 13\) (4) \(3 / 35\) The correct answer is (1), \(1/35\) probability.

Key concepts

Combination Formula

When we wish to select a set of items from a larger group without caring about the order, we use the combination formula. This formula is expressed as \( C(n, r) = \frac{n!}{r!(n-r)!} \). It tells us how many ways we can choose \(r\) items from \(n\) items.

In this exercise, we have numbers from 1 to 15, and we are interested in selecting 3 numbers. This is a classic case of combination.
The formula simplifies the calculation of choices so that we don’t have to list all possibilities. Simply, plug \(n = 15\) and \(r = 3\) into the formula:

• Calculate 15! (factorial of 15)
• Calculate 3! (factorial of 3) and (15-3)! (factorial of 12)
• Divide the factorial of 15 by the product of the factorial of 3 and the factorial of 12

As a result, we find there are 455 possible ways to choose any set of three numbers from this range.

Consecutive Numbers

Consecutive numbers are numbers that follow each other in order, without any gaps. In this context, think of them like steps on a ladder - each step follows sequentially from the last.

For example, the sequence 1, 2, 3 or 3, 4, 5 are consecutive numbers. To solve this exercise, we need to determine how many sets of three consecutive numbers can be formed from the numbers 1 to 15.

When counting consecutive sets:

• Start at the smallest number, here it’s 1, 2, 3.
• Next combination starts at 2, leading to 2, 3, 4.
• Continue this pattern up until the largest starting point, which is 13.

Hence, starting from 1 and moving up to 13, we end up with 13 distinct sets of 3 consecutive numbers from the range 1 to 15.

Probability Calculation

Probability gives us a way to talk about how likely something is to happen. It is calculated using the number of favorable results divided by the number of total possible outcomes.

Let's look at the problem: having found there are 455 possible ways to select 3 numbers, and among these, only 13 sets are consecutive, the favorable outcomes are 13.

In probability terms, we calculate:

• Total outcomes = 455
• Favorable outcomes (consecutive numbers) = 13
• Probability = \( \frac{13}{455} \)

Therefore, the probability that the three chosen numbers are consecutive is \( \frac{13}{455} \). When simplified, this fraction equals \( \frac{1}{35} \). This is how we derive the probability that matches the given answer choices.

Favorable Cases

Favorable cases are those specific scenarios or sets that meet the desired condition in a probability problem. Here, these are the cases where the three chosen numbers are consecutive.

Understanding favorable cases is crucial because they form the numerator in the probability calculation.
In our example, by carefully listing or identifying all sets of three consecutive numbers, we identified that there are 13 such sets.

This meticulous counting can prevent errors:

• Check all possible starting points for triples: 1 to 13.
• Ensure every consecutive set from the range is considered.
• Reaffirm that sets like 13, 14, 15 are included in this count.

Ensuring accurate counting of favorable cases ensures an accurate probability calculation, it’s the cornerstone for finding an effective solution.